Home
Class 10
PHYSICS
A charged particle 'q' is moving with a ...

A charged particle 'q' is moving with a speed 'v' perpendicular to the magnetic field of induction B.Find the radius of the path and time period of the particle.

Text Solution

Verified by Experts

Let us assume that the field is directed into the page as shown in figure. Then the force experienced by the particle F=q v B. we know that this force is always directed perpendicular to velocity . Hence the particle moves along a circular path and the magnetic force on a charged particle acts like a centripetal force.
Let r be the radius of the circular path
We know that centripetal force =`mv^2//r`
`qv B=mv^2//r`
Solving this equation we get , `r=mv //Bq`
Time period of the paricle, `T=2pir//v`
Substituting r in above equation , we get
`T=2pi m // Bq`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • ELECTROMAGNETISM

    VGS PUBLICATION-BRILLIANT|Exercise CREATIVE QUESTION FOR NEW MODEL EXAMINATION (SEC-3)CONCEPTUAL UNDERSTANDING|35 Videos

  • ELECTRIC CURRENT

    VGS PUBLICATION-BRILLIANT|Exercise SECTION - IV (4 Marks Questions)|16 Videos

  • FLOATING BODIES

    VGS PUBLICATION-BRILLIANT|Exercise OBJECTIVE TYPE QUESTIONS|40 Videos

Similar Questions

Explore conceptually related problems

A particle of mass m and charge q is projected into a region having a perpendicular magnetic field B. Find the angle of deviation of the particle as it comes out of the magnetic field if the width d of the region is slightly less than. ( a) (mv)/(qB) " " (b)(mv)/(2qB) " " (c) (2mv)/(qB)

A wire of length 0.1m moves with a speed of 10 ms^(-1) perpendicular to a magnetic field of induction 1Wbm^(-2) . Calculate induced emf

Knowledge Check

  • Orbits of a particle moving in a clrcle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field. The radius of the n^(th) orbital will therefore be proportional to

    A
    `n^(2)`
    B
    n
    C
    `n^(3//2)`
    D
    `n^(1//2)`

  • When a charged particle moving with velocity vecv is subjected to a magnetic field of induction vecB , the force on it is non-zero. This implies that

    A
    and between `vecv` and `vecB` can have any value other than 90°
    B
    angle between `vecv` and `vecB` can have any value other than zero and 180°
    C
    angle between `vecv` and `vecB` is either ero or 180°
    D
    angle between `vecv` and `vecB` is necessarily 90°

  • A charged particle moves through a magnetic field in a direction perpendicular to it. Then the

    A
    Velocity remains unchanged
    B
    Speed of the particle remains unchanged
    C
    Direction of the particle remains unchanged
    D
    Acceleration remains unchanged

    • Similar Questions

    Explore conceptually related problems

    Find the length of the conductor which is moving with a speed of 10 m/s in the direction perpendicular to the direction of magnetic field of induction 0.8T, if it induces an emf of 8V between the ends of the conductor.

    Find the length of the conductor which is moving with a speed of 10 m//s in the direction perpendicular to the direction of magnetic field of induction 0.8T.if it induces an emf of 8V between ends of the conductor.

    (A) : A charged particle is moving in a circle with constant speed in uniform magnetic field. If we increase the speed of particle to twice, its acceleration will become four times. (R) : In circular path with constant speed, acce-leration is given v^(2) /R. If speed is doubled centripetal acceleration will become four times.

    A charged particle of charge q and mass m enters perpendicularly in a magnetic field vecB . Kinetic energy of the particle is E, then frequency of rotation is

    To a charged particle which is moving with a constant initial velocity vecV, uniform magnetic field is applied in the direction of the velocity

    Home
    Profile